6. The length of life of an instrument produced by a machine has a normal ditribution with a mean of 12 months and standard deviation of 2 m

Question

6. The length of life of an instrument produced by a machine has a normal ditribution with a mean of 12 months and standard deviation of 2 months. Find the probability that an instrument produced by this machine will last a) less than 7 months. b) between 7 and 12 months.

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Elliana 2 weeks 2021-10-08T04:10:11+00:00 2 Answers 0

Answers ( )

    0
    2021-10-08T04:11:34+00:00

    Answer:

    Step-by-step explanation:

    Since the length of life of an instrument produced by a machine has a normal distribution, we would apply the formula for normal distribution which is expressed as

    z = (x – µ)/σ

    Where

    x = length of life of instruments in months.

    µ = mean time

    σ = standard deviation

    From the information given,

    µ = 12 months

    σ = 2 months

    a) We want to find the probability that an instrument produced by this machine will last for less than 7 months. It is expressed as

    P(x < 7)

    For x = 7,

    z = (7 – 12)/2 = – 2.5

    Looking at the normal distribution table, the probability corresponding to the z score is 0.0062

    b) between 7 and 12 months is expressed as P(7 ≤ x ≤ 12)

    For x = 7, the probability is 0.0062

    For x = 12,

    z = (12 – 12)/2 = 0

    Looking at the normal distribution table, the probability corresponding to the z score is 0.5

    Therefore,

    P(7 ≤ x ≤ 12) = 0.5 – 0.0062 = 0.4938

    0
    2021-10-08T04:11:55+00:00

    Answer:

    (a) 0.00612.

    (b) 0.49379.

    Step-by-step explanation:

    We have been given that the length of life of an instrument produced by a machine has a normal distribution with a mean of 12 months and standard deviation of 2 months.

    (a) First of all, we will find z-score corresponding to sample score of 7 months as:

    z=\frac{x-\mu}{\sigma}, where,

    z = Z-score,

    x = Sample score,

    \mu = Mean,

    \sigma = Standard deviation.    

    Upon substituting our given values in z-score formula, we will get:

    z=\frac{7-12}{2}=\frac{-5}{2}=-2.5

    Now, we need to find the probability that a z-score is less than -2.5.

    Using normal distribution table, we will get:

    P(z<-2.5)=0.00621

    Therefore, the probability that an instrument produced by this machine will last less than 7 months is 0.00612.

    (b) Let us find z-score corresponding to sample score of 12 months.

    z=\frac{12-12}{2}=\frac{0}{2}=0

    Using formula P(a<z<b)=P(z<b)-P(z<a), we will get:

    P(-2.5<z<0.0)=P(z<0.0)-P(z<-2.5)

    P(-2.5<z<0.0)=0.50000-0.00621

    P(-2.5<z<0.0)=0.49379

    Therefore, the probability that an instrument produced by this machine will last between 7 and 12 months is 0.49379.

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